Two-Dimensional Collisions
When two billiard balls collide at an angle, or a firecracker explodes into fragments flying in every direction, momentum is still conserved — but now we must treat it as a vector quantity in two dimensions. A one-dimensional collision lets us write a single equation; a two-dimensional collision gives us two independent equations, one for each axis. This is not a complication — it is extra information that lets us solve richer, more realistic problems. The fundamental principle is unchanged: in an isolated system (no net external force), the total momentum vector before the event equals the total momentum vector after. Because vectors add component by component, we split every momentum into its x-component and y-component, apply conservation separately to each axis, and then reconstruct magnitudes and directions at the end using the Pythagorean theorem and inverse tangent. Two-dimensional collisions appear everywhere in physics and engineering: subatomic particle tracks in a bubble chamber, the cue-ball break in billiards, spacecraft gravity assists, and car crash reconstruction. Mastering vector momentum conservation is therefore one of the most transferable skills in classical mechanics.
- Set Up a Coordinate System
- Write the Momentum Components Before the Collision
- Apply Conservation Separately on Each Axis
- Solve the Algebraic System
- Reconstruct Magnitude and Direction
Set Up a Coordinate System
Choose convenient x and y axes — often x is along the initial direction of the moving object. Label all objects 1, 2, … and define angles measured from the positive x-axis (or from the initial direction, whichever is clearer). Draw a sketch showing all velocity vectors before and after the collision.
Write the Momentum Components Before the Collision
For each object, compute px = mv·cos θ and py = mv·sin θ. Sum the x-components of all objects to get p_x,total(before), and sum the y-components to get p_y,total(before). Pay careful attention to signs: a vector pointing left or downward has a negative component.
Apply Conservation Separately on Each Axis
Write two equations: Σpx(before) = Σpx(after) and Σpy(before) = Σpy(after). Each equation is an ordinary algebraic equation. If the collision is elastic, a third equation (kinetic energy conservation) is available and must also be satisfied: ΣKE(before) = ΣKE(after). For perfectly inelastic collisions the two objects stick together, so you have a single final-velocity vector to find.
Solve the Algebraic System
With the unknowns (usually one or two final speeds and angles), solve the equations simultaneously. In a typical glancing elastic collision between equal masses where one is initially at rest, the two post-collision velocity vectors are always perpendicular — a beautiful result that follows directly from combining momentum and kinetic energy conservation.
Reconstruct Magnitude and Direction
Once you have final x- and y-components, recover the speed as v = √(vx² + vy²) and the angle as θ = arctan(vy/vx). Always verify that both component equations are satisfied by your answers. Check limiting cases: if the projectile misses, the target should remain at rest and total momentum should equal initial momentum.
The Key Equation
- m_1, m_2 — masses (kg): Masses of object 1 and object 2; they are scalars and appear in both component equations.
- v_{1x,i} — initial x-velocity of object 1 (m/s): The x-component of object 1's velocity before the collision. Obtained via v₁ cos θ₁.
- v_{1y,i} — initial y-velocity of object 1 (m/s): The y-component of object 1's velocity before the collision. Obtained via v₁ sin θ₁.
- v_{1x,f}, v_{1y,f} — final velocity components of object 1 (m/s): The x and y components of object 1's velocity after the collision — often the unknowns to be solved for.
- v_{2x,f}, v_{2y,f} — final velocity components of object 2 (m/s): The x and y components of object 2's velocity after the collision.
Worked Example
Problem. A 2.0 kg puck sliding east at 6.0 m/s strikes a stationary 2.0 kg puck on a frictionless air table. After the collision, puck 1 moves at 30° north of east. The collision is elastic. Find the final speeds of both pucks and the direction of puck 2.
Answer. Puck 1 moves at 3√3 ≈ 5.2 m/s at 30° north of east; puck 2 moves at 3.0 m/s at 60° south of east. The angle between them is exactly 90°.
Origins · 1 minJohn Wallis and how this was discovered
Wallis was among the first to state the principle of conservation of momentum in a rigorous form, publishing his laws of collision (including oblique collisions) in 'Mechanica' (1670–1671). He correctly described how the total quantity of motion — computed as a vector — is preserved in collisions, laying groundwork that Newton would later formalise in the Principia. Wallis was a skilled cryptanalyst who helped the English Parliament decode Royalist messages during the Civil War, all without formal mathematical training in code-breaking.
Related Concepts
- Conservation of Momentum — The entire approach to 2D collisions rests on the law of momentum conservation; understanding the 1D version is the essential prerequisite.
- Elastic and Inelastic Collisions — Whether a 2D collision is elastic or inelastic determines how many equations you have; kinetic energy conservation provides the crucial extra equation for elastic cases.
- Impulse–Momentum Theorem — Impulse is a vector, so a force applied at an angle in 2D changes momentum in that direction — directly connecting impulse to the component method used in 2D collisions.
- Vector Components and Addition — Decomposing velocities into x- and y-components using trigonometry is the fundamental mathematical tool required for every 2D collision problem.